A099741 - OEIS

%I #25 Oct 10 2021 03:41:05

%S 1,1,2,2,2,3,3,3,4,4,4,5,5,5,5,5,5,7,7,7,7,7,7,8,8,8,9,9,9,9,9,9,9,9,

%T 9,12,12,12,12,12,12,12,12,12,12,12,12,13,13,13,13,13,13,16,16,16,16,

%U 16,16,16,16,16,16,16,16,16,16,16,16,16,16,20,20,20,20,20,20,20,20,20,21

%N a(1) = a(2) = 1; a(n) = a([n/2])+a([n/3]) (n >= 3).

%C Let f = f[x,y] be a Fibonacci variant with recurrence f(1) = f(2) = 1; f(n) = f(ceiling((n-1)/x))+f(ceiling((n-2)/y)). This sequence is f[2,3].

%C Nondecreasing. Increases only when n is of the form 2^x*3^y.

%C By the Akra-Bazzi theorem, we have a(n) = Theta(n^e), where e ~ 0.78788491102586978 is the root of the equation (1/2)^e + (1/3)^e = 1. - _Jeffrey Shallit_, Mar 15 2018

%H Robert Israel, <a href="/A099741/b099741.txt">Table of n, a(n) for n = 1..10000</a>

%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Akra%E2%80%93Bazzi_method">Akra-Bazzi method</a>

%F G.f. g(x) satisfies g(x) = x + (1+x)*g(x^2) + (1+x+x^2)*g(x^3). - _Robert Israel_, Mar 15 2018

%e a(19) = a([19/2])+a([19/3]) = a(9)+a(6) = 4+3 = 7.

%p f:= proc(n) option remember; procname(floor(n/2))+procname(floor(n/3)) end proc:

%p f(1):= 1: f(2):= 1:

%p map(f, [$1..100]); # _Robert Israel_, Mar 15 2018

%t a[1] = a[2] = 1;

%t a[n_] := a[n] = a[Floor[n/2]] + a[Floor[n/3]];

%t Array[a, 100] (* _Jean-François Alcover_, Aug 28 2020 *)

%Y Cf. A088468.

%K easy,nonn

%O 1,3

%A _Darrell Minor_, Nov 09 2004

%E Name corrected by _Robert Israel_, Mar 15 2018